The numerical value of the angular velocity of rotation of the earth should be…….. Rad/s in order to make the effective acceleration due to gravity equal to zero.

Explanation

To solve the problem of finding the angular velocity of the Earth's rotation that would make the effective acceleration due to gravity equal to zero, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept**: The effective acceleration due to gravity at the equator when the Earth is rotating is given by the formula:
\[ g_{\text{effective}} = g - R \omega^2 \]
where:
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( R \) is the radius of the Earth (approximately \( 6.371 \times 10^6 \, \text{m} \)),
- \( \omega \) is the angular velocity of the Earth's rotation in radians per second.

2. **Set the Effective Gravity to Zero**: We want to find the angular velocity \( \omega \) such that the effective acceleration due to gravity is zero:
\[ 0 = g - R \omega^2 \]

3. **Rearrange the Equation**: Rearranging the equation gives:
\[ R \omega^2 = g \]

4. **Solve for Angular Velocity**: Dividing both sides by \( R \) gives:
\[ \omega^2 = \frac{g}{R} \]
Taking the square root of both sides results in:
\[ \omega = \sqrt{\frac{g}{R}} \]

5. **Substitute the Values**: Now, substitute the known values of \( g \) and \( R \):
- \( g \approx 9.81 \, \text{m/s}^2 \)
- \( R \approx 6.371 \times 10^6 \, \text{m} \)

Thus, we have:
\[ \omega = \sqrt{\frac{9.81}{6.371 \times 10^6}} \]

6. **Calculate the Value**: Performing the calculation:
\[ \omega = \sqrt{\frac{9.81}{6.371 \times 10^6}} \approx \sqrt{1.540 \times 10^{-6}} \approx 0.00124 \, \text{rad/s} \]

### Final Answer:
The numerical value of the angular velocity of rotation of the Earth should be approximately \( 0.00124 \, \text{rad/s} \) in order to make the effective acceleration due to gravity equal to zero.

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